5.2 Continuous random variables

Transcription

1 5.2 Continuous random variables It is often convenient to think of a random variable as having a whole (continuous) interval for its set of possible values. The devices used to describe continuous probability distributions differ from those that describe discrete probability distributions. Examples of continuous random variables: 24

2 5.2.1 Probability density functions and cumulative distribution functions A probability density function (pdf) is the continuous analogue of a discrete random variable s probability mass function (pmf). Definition A probability density function (pdf) for a continuous random variable X is a nonnegative function f(x) with and such that for all a b, f(x) = 1 P [a X b] = b a f(x)dx f(x) x 25

3 Example 5.17 (Compass needle). Consider a de-magnetized compass needle mounted at its center so that it can spin freely. It is spun clockwise and when it comes to rest the angle, θ, from the vertical, is measured. Let Y = the angle measured after each spin in radians What values can Y take? What form makes sense for f(y)? If this form is adopted, that what must the pdf be? 26

4 Using this pdf, calculate the following probabilities: 1. P [Y < π 2 ] 2. P [ π 2 < Y < 2π] 3. P [ π 6 < Y < π 4 ] 4. P [Y = π 6 ] 27

5 Definition The cumulative distribution function (cdf) of a continuous random variable X is a function F such that F (x) = P [X x] = x f(t)dt F (x) is obtained from f(x) by integration, and applying the fundamental theorem of calculus yields d F (x) = f(x). dx That is, f(x) is obtained from F (x) by differentiation. As with discrete random variables, F has the following properties:

6 Example 5.18 (Compass needle, cont d). Recall the compass needle example, with 1 0 y 2π 2π f(x) = 0 otherwise Find the cdf. For y < 0 For 0 y 2π For y > 2π 29

7 Calculate the following using the cdf: F (1.5) P [Y 4π 5 ] P [Y > 5] P [ π 3 < Y π 2 ] 30

8 5.2.2 Quantiles Recall: Definition The p-quantile of a random variable, X, is the number Q(p) such that P [X Q(p)] = p. In terms of the cumulative distribution function (for a continuous random variable), Example 5.19 (Compass needle, cont d). Recall the compass needle example, with 1 0 y 2π 2π f(x) = 0 otherwise Q(.95): 31

9 You can also calculate quantiles directly from the cdf. Q(.25): 0 y < 0 F (x) = 0 y 2π 1 y 2π 1 otherwise Q(.5) 32

10 5.2.3 Means and variances for continuous distributions It is possible to summarize continuous probability distributions using Definition The mean or expected value of a continuous random variable X is EX = xf(x)dx. Example 5.20 (Compass needle, cont d). Calculate EY where Y is the angle from vertical in radians that a spun needle lands on. 1 0 y 2π 2π f(y) = 0 otherwise 33

11 Example Calculate EX where X follows the following distribution 0 x < 0 f(x) = 1 3 e x/3 x 0 Definition The variance of a continuous random variable X is VarX = (x EX) 2 f(x)dx = x 2 f(x)dx (EX) 2. The standard deviation of X is VarX. 34

12 Example 5.22 (Library books). Let X denote the amount of time for which a book on 2-hour hold reserve at a college library is checked out by a randomly selected student and suppose its density function is 0.5x 0 x 2 f(x) = 0 otherwise Calculate EX and VarX. 35

13 Example 5.23 (Ecology). An ecologist wishes to mark off a circular sampling region having radius 10m. However, the radius of the resulting region is actually a random variable R with pdf 3 f(r) = (10 2 r)2 9 r 11 0 otherwise Calculate ER and SD(R). 36

14 Why does EX 2 = x 2 f(x)dx? Example 5.24 (Ecology, cont d). Calculate the expected area of the circular sampling region. 37

15 For a linear function, g(x) = ax + b, where a and b are constants, E(ax + b) Var(ax + b) Example 5.25 (Ecology, cont d). Calculate the expected value and variance of the diameter of the circular sampling region. 38

16 Definition Standardization is the process of transforming a random variable, X, into the signed number of standard deviations by which it is is above its mean value. Z = X EX SD(X) Z has mean 0 Z has variance (and standard deviation) 1 39

17 5.2.4 A special continuous distribution Just as there are a number of useful discrete distributions commonly applied to engineering problems, there are a number of standard continuous probability distributions. Definition The exponential(α) distribution is a continuous probability distribution with probability density function 1 α f(x) = e x/α x > 0 0 otherwise for α > f(x) α x An Exp(α) random variable measures the waiting time until a specific event that has an equal chance of happening at any point in time. Examples: 40

18 It is straightforward to show for X Exp(α), 1. µ = EX = 0 x 1 α e x/α dx = 2. σ 2 = VarX = (x α) 2 1 α e x/α dx = 0 Further, F (x) has a simple formulation: 41

19 Example 5.26 (Library arrivals, cont d). Recall the example the arrival rate of students at Parks library between 12:00 and 12:10pm early in the week to be about 12.5 students per minute. That translates to a 1/12.5 =.08 minute average waiting time between student arrivals. Consider observing the entrance to Parks library at exactly noon next Tuesday and define the random variable T = the waiting time (min) until the first student passes through the door. Using T Exp(.08), what is the probability of waiting more than 10 seconds (1/6 min) for the first arrival? What is the probability of waiting less than 5 seconds? 42

20 5.2.5 The Normal distribution We have already seen the normal distribution as a bell shaped distribution, but we can formalize this. Definition The normal or Gaussian(µ, σ 2 ) distribution is a continuous probability distribution with probability density f(x) = 1 2πσ 2 e (x µ)2 /2σ 2 for all x for σ > 0. A normal random variable is (often) a finite average of many repeated, independent, identical trials. It is not obvious, but 1. f(x)dx = 1 /2σ 2 2πσ 2 e (x µ)2 dx = 2. EX = x 1 2πσ 2 e (x µ)2 /2σ 2 dx = 3. VarX = (x µ) 2 1 2πσ 2 e (x µ)2 /2σ 2 dx = 43

21 The Calculus I methods of evaluating integrals via anti-differentiation will fail when it comes to normal densities. They do not have anti-derivatives that are expressible in terms of elementary functions. The use of tables for evaluating normal probabilities depends on the following relationship. If X Normal(µ, σ 2 ), P [a X b] = b a 1 2πσ 2 e (x µ)2 /2σ 2 dx = (b µ)/σ (a µ)/σ 1 2π e z2 /2 dz = P [ a µ σ Z b µ ] σ where Z Normal(0, 1). Definition The normal distribution with µ = 0 and σ = 1 is called the standard normal distribution. So, we can find probabilities for all normal distributions by tabulating probabilities for only the standard normal distribution. We will use a table of the standard normal cumulative probability function. Φ(z) = F (z) = z 1 2π e t2 dt. 44

22 Example 5.27 (Standard normal probabilities). P [Z < 1.76] P [.57 < Z < 1.32] We can also do it in reverse, find z such that P [ z < Z < z] =

23 Example 5.28 (Baby food). J. Fisher, in his article Computer Assisted Net Weight Control (Quality Progress, June 1983), discusses the filling of food containers with strained plums and tapioca by weight. The mean of the values portrayed is about 137.2g, the standard deviation is about 1.6g, and data look bell-shaped. Let W = the next fill weight. Let s find the probability that the next jar contains less food by mass than it s supposed to (declared weight = g). 46

24 47

25 Example 5.29 (More normal probabilities). Using the standard normal table, calculate the following: P (X 3), X Normal(2, 64) P (X > 7), X Normal(6, 9) P ( X 1 > 0.5), X Normal(2, 4) 48

26 We can find standard normal quantiles by using the standard normal table in reverse. Example 5.30 (Baby food, cont d). For the jar weights X Normal(137.2, ), find Q(0.1). 49

27 Example 5.31 (Normal quantiles). Find: Q(0.95) of X Normal(9, 3). c such that P ( X 2 > c) = 0.01, X Normal(2, 4) 50

28 Table B.3 Standard Normal Cumulative Probabilities (z) = z ( ) 1 exp t2 dt 2π 2 z

29 Table B.3 Standard Normal Cumulative Probabilities (continued) z This table was generated using MINITAB. 789